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To solve this problem, we need to verify whether the Mean Value Theorem (MVT) applies to the given functions over their specified intervals. The MVT requires that:

1. The function is continuous on the closed interval \([a, b]\).
2. The function is differentiable on the open interval \((a, b)\).

If the hypotheses are satisfied, the theorem guarantees that there exists some \(c \in (a, b)\) such that:

\[
f'(c) = \frac{f(b) - f(a)}{b - a}
\]

Now let's work through each function:

### (a) \( f(x) = 1 - x^{\frac{2}{3}} \) on the interval \([-1, 1]\)
1. **Continuity**: The function \(f(x) = 1 - x^{\frac{2}{3}}\) is continuous on \([-1, 1]\) because \(x^{\frac{2}{3}}\) is defined for all \(x \in \mathbb{R}\).
   
2. **Differentiability**: To check differentiability, compute the derivative:

   \[
   f'(x) = -\frac{2}{3}x^{-\frac{1}{3}}
   \]

   The derivative is not defined at \(x = 0\) because \(x^{-\frac{1}{3}}\) becomes undefined at \(x = 0\). Hence, the function is **not differentiable** on the open interval \((-1, 1)\), and the Mean Value Theorem **does not apply** in this case.

### (b) \( f(x) = x^3 - 3x + 4 \) on the interval \([-2, 2]\)
1. **Continuity**: The function is a polynomial, so it is continuous on \([-2, 2]\).
   
2. **Differentiability**: The function is also differentiable everywhere since polynomials are differentiable for all \(x \in \mathbb{R}\). Let's compute the derivative:

   \[
   f'(x) = 3x^2 - 3
   \]

   Since the function satisfies the hypotheses of the Mean Value Theorem, we can proceed to find \(c\).

3. **Mean Value Theorem**: The MVT guarantees that there exists some \(c \in (-2, 2)\) such that:

   \[
   f'(c) = \frac{f(2) - f(-2)}{2 - (-2)}
   \]

   First, calculate \(f(2)\) and \(f(-2)\):

   \[
   f(2) = 2^3 - 3(2) + 4 = 8 - 6 + 4 = 6
   \]
   \[
   f(-2) = (-2)^3 - 3(-2) + 4 = -8 + 6 + 4 = 2
   \]

   Therefore, the right-hand side of the MVT equation is:

   \[
   \frac{f(2) - f(-2)}{4} = \frac{6 - 2}{4} = 1
   \]

   Now, set \(f'(c) = 1\):

   \[
   3c^2 - 3 = 1
   \]
   \[
   3c^2 = 4 \quad \Rightarrow \quad c^2 = \frac{4}{3}
   \quad \Rightarrow \quad c = \pm \frac{2}{\sqrt{3}} = \pm \frac{2\sqrt{3}}{3}
   \]

Thus, the values of \(c\) that satisfy the Mean Value Theorem for part (b) are:

\[
c = \pm \frac{2\sqrt{3}}{3}
\]

---

Would you like more details or clarifications? Here are five related questions you might find useful:

1. How do you determine whether a function is continuous?
2. Why does the Mean Value Theorem require differentiability?
3. Can you apply the Mean Value Theorem to piecewise functions?
4. What are the geometric implications of the Mean Value Theorem?
5. How does the Mean Value Theorem relate to Rolle's Theorem?

**Tip:** Always check the differentiability of a function on the open interval when using the Mean Value Theorem, even if it is continuous.

Does the function satisfy the hypotheses of the Mean Value Theorem on the given interval? If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem, otherwise check whether such a number c exists or not. (a) f(x) = 1 - x^(2/3), [-1,1] (b) f(x) = x^3 - 3x + 4, [-2,2]

Explore whether the functions f(x) = 1 - x^(2/3) and f(x) = x^3 - 3x + 4 satisfy the Mean Value Theorem. Includes detailed steps on checking continuity, differentiability, and finding the values of c for which f'(c) = (f(b) - f(a)) / (b - a).

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